Đáp án: D
Giải thích các bước giải:
$\begin{array}{l}
{x^3} + xy + \left( {2x + y} \right) = 2{y^3} + 2xy\left( {x + 2y} \right)\\
\Leftrightarrow {x^3} + 2{x^2}y + x{y^2} = 2{y^3} + 2{x^2}y + 4x{y^2}\\
\Leftrightarrow {x^3} - 3x{y^2} - 2{y^3} = 0\\
\Leftrightarrow \left( {{x^3} - x{y^2}} \right) - \left( {2x{y^2} + 2{y^3}} \right) = 0\\
\Leftrightarrow x\left( {{x^2} - {y^2}} \right) - 2{y^2}\left( {x + y} \right) = 0\\
\Leftrightarrow \left( {x + y} \right)\left( {{x^2} - xy - 2{y^2}} \right) = 0\\
\Leftrightarrow \left( {x + y} \right)\left( {x + y} \right)\left( {x - 2y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - y\\
x = 2y
\end{array} \right.\\
Do:x,y > 0 \Rightarrow x = 2y\\
\Rightarrow \log _3^2\left( {\frac{{{x^2}}}{x}} \right) - m{\log _3}\left( {\frac{{{x^2}}}{x}} \right) + 2m - 4 = 0\\
\Leftrightarrow \log _3^2x - m{\log _3}x + 2m - 4 = 0\\
Dat:{\log _3}x = t\left( {x \in \left[ {1;3} \right] \Rightarrow t \in \left[ {0;1} \right]} \right)\\
\Rightarrow {t^2} - mt + 2m - 4 = 0\\
\Leftrightarrow \left( {t - 2} \right)\left( {t + 2} \right) - m\left( {t - 2} \right) = 0\\
\Leftrightarrow \left( {t - 2} \right)\left( {t + 2 - m} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 2\left( {ktm} \right)\\
t = m - 2
\end{array} \right.\\
\Rightarrow 0 \le m - 2 \le 1\\
\Rightarrow 2 \le m \le 3
\end{array}$
